A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
A 661N force is being applied to a 121kg block. The coefficient of static friction between the block and the surface is 0.410. What is the minimal amount of additional mass that must be added on top of the block to prevent the block from moving?
 2 years ago
A 661N force is being applied to a 121kg block. The coefficient of static friction between the block and the surface is 0.410. What is the minimal amount of additional mass that must be added on top of the block to prevent the block from moving?

This Question is Closed

BTaylor
 2 years ago
Best ResponseYou've already chosen the best response.0The frictional force is the coefficient of friction times the normal force. Here, you want the frictional force to equal 661, so there is no movement. So:\[F_{fr} = \mu_s \times F_N \rightarrow 661 = 0.410 \times mg\] From this, you can solve for mg, and you get mg = 1612 dividing by g (9.8), you get m = 164.5. So, since you already have 121 kg, you only need 43.5 more.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.